Answer
Center: $(-4,-2)$ $;$ Radius: $2$
Work Step by Step
$x^{2}+y^{2}+8x+4y+16=0$
Take $16$ to the right side:
$x^{2}+y^{2}+8x+4y=-16$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+8x)+(y^{2}+4y)=-16$
Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside each parentheses and also on the right side of the equation.
For the first group, $b=8$ and for the second group, $b=4$
$\Big[x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}\Big]+\Big[y^{2}+4y+\Big(\dfrac{4}{2}\Big)^{2}\Big]=-16+\Big(\dfrac{8}{2}\Big)^{2}+\Big(\dfrac{4}{2}\Big)^{2}$
$(x^{2}+8x+16)+(y^{2}+4y+4)=-16+16+4$
$(x^{2}+8x+16)+(y^{2}+4y+4)=4$
Factor the expressions inside each parentheses, which are perfect square trinomials:
$(x+4)^{2}+(y+2)^{2}=4$
This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$
It can be seen from the obtained equation that $(h,k)=(-4,-2)$, so the center of the circle is the point $(-4,-2)$
The radius is:
$r^{2}=4$
$\sqrt{r^{2}}=\sqrt{4}$
$r=2$
The graph is: